Therefore c x y ax b y lizhi wang lzwangiastateedu ie

Therefore, c
>
x ≤ y
>
Ax ≤ b
>
y .
Lizhi Wang ([email protected]) IE 534 Linear Programming September 12, 2012 11 / 21

Outline
1
Review
2
Complementary slackness
3
Weak duality
4
Strong duality
5
Farkas lemma
6
Primal-dual possibilities
Lizhi Wang ([email protected]) IE 534 Linear Programming September 12, 2012 12 / 21

Strong duality
Theorem (Strong duality)
If x and y are optimal solutions to max { c
>
x : Ax ≤ b, x ≥ 0 } and
min { b
>
y : A
>
y ≥ c, y ≥ 0 } , respectively, then c
>
x = b
>
y.
Proof.
Since x and y are optimal, by complementary slackness, we have
0 = ( b - Ax )
>
y = b
>
y - ( Ax )
>
y = b
>
y - y
>
Ax,
and
0 = ( A
>
y - c )
>
x = y
>
Ax - c
>
x.
Therefore, c
>
x = y
>
Ax = b
>
y .
Lizhi Wang ([email protected]) IE 534 Linear Programming September 12, 2012 13 / 21

Outline
1
Review
2
Complementary slackness
3
Weak duality
4
Strong duality
5
Farkas lemma
6
Primal-dual possibilities
Lizhi Wang ([email protected]) IE 534 Linear Programming September 12, 2012 14 / 21

Rethink unboundedness
Consider the standard from LP max
x
{ c
>
x : Ax ≤ b, x ≥ 0 } .
An LP being unbounded means two things
1
a feasible solution x
0
, which satisfies Ax
0
≤ b, x
0
≥ 0
2
a direction Δ x , which leads lim
λ →∞
c
>
( x
0
+ λ Δ x ) → ∞ without
making ( x
0
+ λ Δ x ) infeasible
This direction Δ x is called an extreme ray . It satisfies:
A Δ x ≤ 0 , Δ x ≥ 0 , and c
>
Δ x > 0 .
Consider the dual LP min
y
{ b
>
y : A
>
y ≥ c, y ≥ 0 } .
The dual extreme ray Δ y satisfies:
A
>
Δ y ≥ 0 , Δ y ≥ 0 , and b
>
Δ y < 0 .
Lizhi Wang ([email protected]) IE 534 Linear Programming September 12, 2012 15 / 21

Farkas lemma
Theorem (Farkas lemma)
Let A ∈ R
m × n
and b ∈ R
m × 1
be a matrix and a vector, respectively.
Then exactly one of the following two alternatives holds:
(a) There exists some x ≥ 0 such that Ax ≤ b .
(b) There exists some y ≥ 0 such that A
>
y ≥ 0 and b
>
y < 0 .
Proof.
(a) true ⇒ (b) false: If (a) is true, then for any y ≥ 0 such that A
>
y ≥ 0 ,
we have b
>
y ≥ ( Ax )
>
y = x
>
A
>
y ≥ 0 , which means that (b) is false.